{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["-1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\sqrt[3]{{{x}^{3}}+9{{x}^{2}}}=x+3 $<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_}","hint":" L\u1eadp ph\u01b0\u01a1ng hai v\u1ebf ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: L\u00e2p ph\u01b0\u01a1ng hai v\u1ebf <br\/> B\u01b0\u1edbc 2: Bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $ax=b$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> Ta c\u00f3:<br\/>$\\begin{aligned} & \\,\\,\\,\\sqrt[3]{{{x}^{3}}+9{{x}^{2}}}=x+3 \\\\ & \\Leftrightarrow {{x}^{3}}+9{{x}^{2}}\\,={{x}^{3}}+9{{x}^{2}}+27x+27 \\\\ & \\Leftrightarrow 27x+27=0 \\\\ & \\Leftrightarrow x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=-1 \\\\ \\end{aligned}$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-1\\}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $-1$<\/span> <\/span><\/span>"}]}],"id_ques":701},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["0"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\sqrt[3]{1-x}+\\sqrt[3]{1+x}=2$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_}","hint":" L\u1eadp ph\u01b0\u01a1ng hai v\u1ebf v\u00e0 s\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c <br\/>${{\\left( a+b \\right)}^{3}}= a^3 + 3a^2b + 3ab^2 + b^3 ={{a}^{3}}+{{b}^{3}}+3ab\\left( a+b \\right)$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: L\u1eadp ph\u01b0\u01a1ng hai v\u1ebf v\u00e0 s\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c <br\/> ${{\\left( a+b \\right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\\left( a+b \\right)$ <br\/>B\u01b0\u1edbc 2: Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/> $ \\,\\,\\,\\,\\,\\,\\,\\sqrt[3]{1-x}+\\sqrt[3]{1+x}=2 $<br\/>$\\Leftrightarrow 1-x+1+x+\\,$$3\\,\\sqrt[3]{\\left( 1-x \\right)\\left( 1+x \\right)}\\,$$.\\left( \\,\\sqrt[3]{1-x}+\\sqrt[3]{1+x} \\right)=8 $<br\/>$ \\Leftrightarrow \\sqrt[3]{\\left( 1-x \\right)\\left( 1+x \\right)}$$.\\underbrace{\\left( \\,\\sqrt[3]{1-x}+\\sqrt[3]{1+x} \\right)}_{=2}=2 $<br\/>$\\begin{aligned}& \\Leftrightarrow \\sqrt[3]{\\left( 1-x \\right)\\left( 1+x \\right)}=1 \\\\ & \\Leftrightarrow 1-{{x}^{2}}=1 \\\\ & \\Leftrightarrow {{x}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,=0 \\\\ & \\Leftrightarrow x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=0 \\\\ \\end{aligned}$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{0\\}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $0$<\/span><\/span> <\/span>"}]}],"id_ques":702},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["9"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $E=\\sqrt[3]{9.\\left( 23+8\\sqrt{7} \\right).\\left( 23-8\\sqrt{7} \\right)}$ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $E =\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$","hint":" \u0110\u01b0a $23\\pm 8\\sqrt{7}$ v\u1ec1 d\u1ea1ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(a \\pm b)^2$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: \u0110\u01b0a bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng $(a \\pm b)^2$<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng: $(a-b)(a+b)=a^2 -b^2$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\begin{aligned} E&=\\sqrt[3]{9.\\left( 23+8\\sqrt{7} \\right).\\left( 23-8\\sqrt{7} \\right)} \\\\ & =\\sqrt[3]{9.\\left( {{4}^{2}}+2.4.\\sqrt{7}+\\sqrt{{{7}^{2}}} \\right).\\left( {{4}^{2}}-2.4.\\sqrt{7}+\\sqrt{{{7}^{2}}} \\right)} \\\\ & =\\sqrt[3]{9.{{\\left( 4+\\sqrt{7} \\right)}^{2}}.{{\\left( 4-\\sqrt{7} \\right)}^{2}}} \\\\ & =\\sqrt[3]{9.{{\\left( 16-7 \\right)}^{2}}} \\\\ & =9 \\\\ \\end{aligned}$ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $9$<\/span><br\/><b>Nh\u1eadn x\u00e9t:<\/b> Ngo\u00e0i ra, v\u1edbi b\u00e0i to\u00e1n n\u00e0y ta c\u00f3 th\u1ec3 s\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$ ngay t\u1eeb b\u01b0\u1edbc \u0111\u1ea7u ti\u00ean:<br\/>Ta c\u00f3: <br\/>$\\begin{align} & E=\\sqrt[3]{9.(23+8\\sqrt{7})(23-8\\sqrt{7})} \\\\ & \\,\\,\\,\\,\\,\\,=\\sqrt[3]{9.\\left[ {{23}^{2}}-{{\\left( 8\\sqrt{7} \\right)}^{2}} \\right]} \\\\ & \\,\\,\\,\\,\\,\\,=\\sqrt[3]{9.(529-448)} \\\\ & \\,\\,\\,\\,\\,\\,=\\sqrt[3]{9.81} \\\\ & \\,\\,\\,\\,\\,\\,=\\sqrt[3]{{{9}^{3}}} \\\\ & \\,\\,\\,\\,\\,\\,=9 \\\\ \\end{align}$<\/span><\/span>"}]}],"id_ques":703},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["-4"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"T\u00ednh gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c<br\/>$B=\\dfrac{a+1}{\\sqrt[3]{{{a}^{2}}}-\\sqrt[3]{a}+1}$ v\u1edbi $\\,a= -125$<br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $B=$ _input_","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Ph\u00e2n t\u00edch t\u1eed th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed. <br\/>B\u01b0\u1edbc 2: R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/>B\u01b0\u1edbc 3: Thay $a$ v\u00e0o $B$ \u0111\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{align} B&=\\dfrac{a+1}{\\sqrt[3]{{{a}^{2}}}-\\sqrt[3]{a}+1} \\\\ & =\\dfrac{\\sqrt[3]{{{a}^{3}}}+1}{\\sqrt[3]{{{a}^{2}}}-\\sqrt[3]{a}+1} \\\\ & =\\dfrac{\\left( \\sqrt[3]{a}+1 \\right)\\left( \\sqrt[3]{{{a}^{2}}}-\\sqrt[3]{a}+1 \\right)}{\\sqrt[3]{{{a}^{2}}}-\\sqrt[3]{a}+1} \\\\ & =\\sqrt[3]{a}+1 \\\\ \\end{align}$<br\/>Thay $a= -125$ v\u00e0o bi\u1ec3u th\u1ee9c $B$, ta \u0111\u01b0\u1ee3c: <br\/>$B=\\sqrt[3]{-125}+1=-5+1=-4$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $-4$<\/span><\/span> <\/span>"}]}],"id_ques":704},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/>$\\sqrt[3]{{{a}^{3}}-3{{a}^{2}}+3a-1}+\\sqrt[3]{27-27a+9{{a}^{2}}-{{a}^{3}}}=$_input_","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng $(a \\pm b)^3$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c trong c\u0103n v\u1ec1 d\u1ea1ng $(a \\pm b)^3$<br\/>B\u01b0\u1edbc 2: C\u1ed9ng tr\u1eeb \u0111\u01a1n th\u1ee9c \u0111\u1ed3ng d\u1ea1ng<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\sqrt[3]{{{a}^{3}}-3{{a}^{2}}+3a-1}\\,$$+\\sqrt[3]{27-27a+9{{a}^{2}}-{{a}^{3}}} $<br\/>$ =\\sqrt[3]{{{\\left( a-1 \\right)}^{3}}}+\\sqrt[3]{{{\\left( 3-a \\right)}^{3}}}$<br\/>$ =a-1+3-a$<br\/>$ =2$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 2 <\/span><\/span>"}]}],"id_ques":705},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank_random","correct":[[["3"],["1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop9/toan/daiso/bai6/lv3/img\/9.jpg' \/><center> T\u00ecm $a,b$ nguy\u00ean \u0111\u1ec3 $\\sqrt[3]{6\\sqrt{3}-10}=\\sqrt{a}-b $ <br\/> <b> \u0110\u00e1p \u00e1n: <\/b> $a=$_input_; $b=$_input_ <\/span>","hint":"\u00c1p d\u1ee5ng $a^3-3a^2b+3ab^2-b^3=(a-b)^3$ ","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & \\,\\,\\,\\,\\sqrt[3]{6\\sqrt{3}-10} \\\\ & =\\sqrt[3]{\\sqrt{3^3}-3.\\sqrt{3^2}.1+3.\\sqrt{3}.1-1}\\\\ & =\\sqrt[3]{{{\\left( \\sqrt{3}-1 \\right)}^{3}}} \\\\ & =\\sqrt{3}-1 \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $3$ v\u00e0 $1$ <\/span><\/span> "}]}],"id_ques":706},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/>$\\dfrac{\\sqrt[3]{6\\sqrt{3}-10}}{\\sqrt{4-2\\sqrt{3}}}=$_input_","hint":"\u00c1p d\u1ee5ng Quy t\u1eafc \u0111\u01b0a th\u1eeba s\u1ed1 ra ngo\u00e0i d\u1ea5u c\u0103n. ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $ \\sqrt[3]{6\\sqrt{3}-10}=\\sqrt[3]{(\\sqrt{3}- 1)^3}$<br\/>Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $\\sqrt{4-2\\sqrt{3}}=\\sqrt{{{\\left( \\sqrt{3}-1 \\right)}^{2}}}$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>$\\begin{align} & \\,\\,\\,\\,\\dfrac{\\sqrt[3]{6\\sqrt{3}-10}}{\\sqrt{4-2\\sqrt{3}}} \\\\ & =\\dfrac{\\sqrt[3]{\\sqrt{{{3}^{3}}}-3.\\sqrt{{{3}^{2}}}.1+3.\\sqrt{3}.1-{{1}^{3}}}\\,\\,}{\\sqrt{{{\\left( \\sqrt{3}-1 \\right)}^{2}}}} \\\\ & =\\dfrac{\\sqrt[3]{{{\\left( \\sqrt{3}-1 \\right)}^{3}}}}{\\sqrt{3}-1} \\\\ & =\\dfrac{\\sqrt{3}-1}{\\sqrt{3}-1}=1 \\\\ \\end{align}$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $1$ <\/span><\/span> <\/span>"}]}],"id_ques":707},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng ","title_trans":"","temp":"fill_the_blank","correct":[[["5"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","type_check":"","ques":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c <br\/>$ B=\\left( \\sqrt[3]{4}+\\sqrt[3]{6}+\\sqrt[3]{9} \\right)\\,$$\\left( \\sqrt[3]{4}-\\sqrt[3]{6}+\\sqrt[3]{9} \\right)\\,$$\\left( \\sqrt[3]{2}+\\sqrt[3]{3} \\right)\\left( \\sqrt[3]{3}-\\sqrt[3]{2} \\right)$<br\/>\u0110\u00e1p \u00e1n: <\/b> $B=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{} $ ","hint":"Nh\u00f3m c\u00e1c th\u1eeba s\u1ed1 \u0111\u1ec3 xu\u1ea5t hi\u1ec7n h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^3\\pm b^3$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$ B=\\left( \\sqrt[3]{4}+\\sqrt[3]{6}+\\sqrt[3]{9} \\right)\\,$$\\left( \\sqrt[3]{4}-\\sqrt[3]{6}+\\sqrt[3]{9} \\right)\\,$$\\left( \\sqrt[3]{2}+\\sqrt[3]{3} \\right)\\left( \\sqrt[3]{3}-\\sqrt[3]{2} \\right) $<br\/>$ =\\left[ \\left( \\sqrt[3]{4}+\\sqrt[3]{6}+\\sqrt[3]{9} \\right)\\left( \\sqrt[3]{3}-\\sqrt[3]{2} \\right) \\right]\\,$$\\left[ \\left( \\sqrt[3]{4}-\\sqrt[3]{6}+\\sqrt[3]{9} \\right)\\left( \\sqrt[3]{2}+\\sqrt[3]{3} \\right) \\right]$<br\/>$ =\\left( \\sqrt[3]{{{3}^{3}}}-\\sqrt[3]{{{2}^{3}}} \\right)\\left( \\sqrt[3]{{{2}^{3}}}+\\sqrt[3]{{{3}^{3}}} \\right) $<br\/>$ =\\left( 3-2 \\right)\\left( 3+2 \\right)$<br\/>$=5 $ <br\/> <span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $5$<\/span><\/span>"}]}],"id_ques":708},{"time":24,"part":[{"title":"H\u00e3y ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang ","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":10,"ques":"Tr\u1ee5c c\u0103n th\u1ee9c \u1edf m\u1eabu $ \\dfrac{1}{\\sqrt[3]{2}-1}$ \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3 l\u00e0: ","select":["A. $\\dfrac{\\sqrt[3]{4}+\\sqrt[3]{2}+\\sqrt[3]{1}}{3}$ ","B. $\\dfrac{\\sqrt[3]{4}-\\sqrt[3]{2}+\\sqrt[3]{1}}{3}$","C. $\\sqrt[3]{4}-\\sqrt[3]{2}+\\sqrt[3]{1}$","D. $\\sqrt[3]{4}+\\sqrt[3]{2}+\\sqrt[3]{1}$"],"hint":"Nh\u00e2n c\u1ea3 t\u1eed v\u00e0 m\u1eabu v\u1edbi bi\u1ec3u th\u1ee9c li\u00ean h\u1ee3p c\u1ee7a m\u1eabu","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>Nh\u00e2n c\u1ea3 t\u1eed v\u1edbi m\u1eabu v\u1edbi bi\u1ec3u th\u1ee9c $\\sqrt[3]{4}+\\sqrt[3]{2}+\\sqrt[3]{1}$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<br\/><\/span>Ta c\u00f3:<br\/>$\\begin{align}\\,\\,\\,\\,& \\dfrac{1}{\\sqrt[3]{2}-1}\\\\& =\\dfrac{\\sqrt[3]{{{2}^{2}}}+\\sqrt[3]{2}+\\sqrt[3]{{{1}^{2}}}}{\\left( \\sqrt[3]{2}-1 \\right)\\left( \\sqrt[3]{{{2}^{2}}}+\\sqrt[3]{2}+\\sqrt[3]{{{1}^{2}}} \\right)} \\\\ & =\\dfrac{\\sqrt[3]{4}+\\sqrt[3]{2}+\\sqrt[3]{1}}{\\sqrt[3]{{{2}^{3}}}-1} \\\\ & =\\sqrt[3]{4}+\\sqrt[3]{2}+\\sqrt[3]{1} \\\\ \\end{align}$ <br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span>","column":2}]}],"id_ques":709},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"V\u1edbi $ab\\ne 0$ ta c\u00f3 $\\dfrac{\\sqrt[3]{{{a}^{5}}{{b}^{7}}}}{\\sqrt[3]{{{a}^{2}}b}}-\\dfrac{\\sqrt[3]{{{a}^{4}}{{b}^{8}}}}{\\sqrt[3]{a{{b}^{2}}}}$=0 ","select":["A. \u0110\u00fang ","B. Sai"],"hint":"V\u1edbi b$\\ne$ 0 ta c\u00f3 $\\dfrac{\\sqrt[3]{a}}{\\sqrt[3]{b}}=\\sqrt[3]{\\dfrac{a}{b}}$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<br\/><\/span>B\u01b0\u1edbc 1: V\u1edbi b$\\ne$ 0 ta c\u00f3 $\\dfrac{\\sqrt[3]{a}}{\\sqrt[3]{b}}=\\sqrt[3]{\\dfrac{a}{b}}$<br\/>B\u01b0\u1edbc 2: R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<br\/><\/span>$\\dfrac{\\sqrt[3]{{{a}^{5}}{{b}^{7}}}}{\\sqrt[3]{{{a}^{2}}b}}-\\dfrac{\\sqrt[3]{{{a}^{4}}{{b}^{8}}}}{\\sqrt[3]{a{{b}^{2}}}}\\,$$=\\sqrt[3]{\\dfrac{{{a}^{5}}{{b}^{7}}}{{{a}^{2}}b}}-\\sqrt[3]{\\dfrac{{{a}^{4}}{{b}^{8}}}{a{{b}^{2}}}}\\,$$=\\sqrt[3]{{{a}^{3}}{{b}^{6}}}-\\sqrt[3]{{{a}^{3}}{{b}^{6}}}=0$ <br\/> Do \u0111\u00f3 kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":710}],"lesson":{"save":0,"level":3}}